Problem: Graph this system of equations and solve. $x-4y = 4$ $6x+8y = -40$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: Convert the first equation, $x-4y = 4$ , to slope-intercept form. $y = \dfrac{1}{4} x - 1$ The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $\dfrac{1}{4}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $4$ positions to the right. $4$ positions to the right. Graph the blue line so it passes through $(0, -1)$ and $(4, 0)$ Convert the second equation, $6x+8y = -40$ , to slope-intercept form. $y = -\dfrac{3}{4} x - 5$ The y-intercept for the second equation is $-5$ , so the second line must pass through the point $(0, -5)$ The slope for the second equation is $-\dfrac{3}{4}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) You must also move $4$ positions to the right. $4$ positions to the right. $3$ positions down from $(0, -5)$ is $(4, -8)$ Graph the green line so it passes through $(0, -5)$ and $(4, -8)$ The solution is the point where the two lines intersect. The lines intersect at $(-4, -2)$.